Jeremiah makes $25\%$ of the three-point shots he attempts. For a warm up, Jeremiah likes to shoot three-point shots until he makes one. Let $M$ be the number of shots it takes Jeremiah to make his first three-point shot. Assume that the results of each shot are independent. Find the probability that it takes Jeremiah fewer than $4$ attempts to make his first shot. You may round your answer to the nearest hundredth. $P(M<4)=$
Without a fancy calculator On each shot: $P({\text{make}})=0.25$ $P(\text{miss}})=0.75$ If it takes Jeremiah fewer than $4$ attempts to make his first shot, here are the possible sequences of shots: make miss, make miss, miss, make We can find the probability of each sequence and add those probabilities together. $\begin{aligned} P({\text{make}})&=0.25 \\\\\\ P(\text{miss}}, {\text{make}})&=(0.75})({0.25})\\\\&=0.1875 \\\\\\ P(\text{miss}}, \text{miss}}, {\text{make}} )&=(0.75})(0.75})({0.25}) \\\\&=0.140625 \\\\\\ P(M<4)&=0.25+0.1875+0.140625 \\\\&=0.578125 \end{aligned}$ [Is there another way?] $P(M<4)=0.578125 \approx 0.58$